Question

Thirty small communities in Connecticut (population near 10,000 each) gave an average of 138.5 reported cases of larceny per year. Assume that is known to be 42.1 cases per year. (a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error (c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error (d) Compare the margins of error for parts () through (c). As the confidence levels increase, do the margins of error increase? As the confidence level increases, the margin of error remains the same. As the confidence level increases, the margin of error decreases. O As the confidence level increases, the margin of error increases (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length? O As the confidence level increases, the confidence interval decreases in length o As the confidence level increases, the confidence interval increases in length. O As the confidence level increases, the confidence interval remains the same length Need Help? Red 4.-/2 points understatt2 7.2.001. ER What are the degrees of freedom for Student's t distribution when the sample size is 22? dl. Use the Student's distribution to find for a 0.95 confidence level when the sample is 22. (Round your answer to three decimal places.) Need Help? Red

Answer 1

We have the known population standard deviation so we will use standard normal distribution

$n=10000\;\bar{x}=138.5\;\sigma=42.1$

The (1 - $\alpha$) confidence interval for population mean with known population SD is

$(\bar{x}\bar{+}MOE)$

$(138.5\bar{+}MOE)$

MOE (margin of error) = Critical value * Standard error

= $Z_{\alpha/2}$ * $\frac{42.1}{\sqrt{10000}}$

Where $Z_{\alpha/2}$ can be found using normal distribution tables or excel function 'normsinv'

Confidence levels	(a) 90%	(b) 95%	(c) 99%
$\alpha$	0.10	0.05	0.01
$\alpha$/2	0.05	0.025	0.005
Critical value	1.6449	1.96	2.5758
Margin of error (the margin of error increases with increase in $\alpha$)	0.00692	0.00825	0.01084
Lower limit	138.4931	138.4918	138.4892
Upper limit	138.5069	138.5082	138.5108
Width (length) (Width is increasing)	0.01385	0.01650	0.02169

(d) As the confidence level increases the margin of error increase (option 3)

This is because increasing the level allows for more accuracy and makes it stricter. Thus it incorporates more values so that the possibility of the true population mean lying in the interval is increased.

(e) As the confidence level increases the confidence level length increase (option 2)

This is because increasing the level allows for more accuracy and makes it stricter. Thus it incorporates more values so that the possibility of the true population mean lying in the interval is increased.

t-dist

df = n - 1 = 22 - 1

df = 21

The $\alpha$ = 1 - 0.95 = 0.05

$t_{\alpha/2,df}=t_{0.025,21}$

$\mathbf{t_{c}=2.0796=2.08}$ (using t-dist tables)